 Homework 9 Solutions 1 Chapter 14 Enrique Trevi˜

```Homework 9 Solutions
Enrique Trevi˜
no
November 20, 2014
1
Chapter 14
Problem 1. (Exercise 2)
Compute all Xg and all Gx for each of the following permutation groups.
(a) X = {1, 2, 3},
G = S3 = {(1), (12), (13), (23), (123), (132)}
(b) X = {1, 2, 3, 4, 5, 6},
G = {(1), (12), (345), (354), (12)(345), (12)(354)}
Solution 1. To turn in.
Problem 2. (Exercise 3)
Compute the G-equivalence classes of X for each of the G-sets in the previous Exercise. For each x ∈ X
verify that |G| = |Ox | · |Gx |.
Solution 2. To turn in.
Problem 3. (Exercise 4)
Let G be the additive group of real numbers. Let the action of θ ∈ G on the real plane R2 be given by
rotating the plane counterclockwise about the origin through θ radians. Let P be a point on the plane other
than the origin.
(a) Show that R2 is a G-set.
(b) Describe geometrically the orbit containing P .
(c) Find the group GP .
Solution 3. To turn in.
Problem 4. (Exercise 5)
Let G = A4 and suppose that G acts on itself by conjugation; that is, (g, h) 7→ ghg −1 .
(a) Determine the conjugacy classes (orbits) of each element of G.
(b) Determine all of the isotropy subgroups for each element of G.
Solution 4. To turn in.
Problem 5. (Exercise 6)
Find the conjugacy classes and the class equation for each of the following groups.
(a) S4
(b) D5
(c) Z9
1
(d) Q8
Solution 5.
(a)
S4 = {(1), (12), (13), (14), (23), (24), (34), (123), (132), (124), (142), (134), (143), (234), (243),
(12)(34), (13)(24), (14)(23), (1234), (1243), (1324), (1342), (1423), (1432)}.
Z(S4 ) = {(1)}.
We know that if σ ∈ S4 , then σ(12)σ −1 = (σ(1), σ(2)), so that makes it easier to calculate the conjugacy
class of (12). For example
(1342)(12)(1342)−1 = ((1342)(1), (1342)(2)) = (31) = (13).
The orbit of (12) is
O(12) = {(12), (23), (24), (13), (14), (34)}.
It turned out to be all transpositions. The orbit of (123) is
O(123) = {(123), (132), (124), (142), (134), (143), (234), (243)}.
The orbit of (12)(34) is
O(12)(34) = {(12)(34), (13)(24), (14)(23)}.
There is one more conjugacy class:
O(1234) = {(1234), (1243), (1324), (1342), (1423), (1432)}.
The conjugacy classes break out in cycle types.
|S4 | = 24,
|Z(G)| = 1,
|O(12) | = 6,
|O(123) | = 8,
|O(12)(34) | = 3,
|O(1234) = 6|,
so
24 = 1 + 6 + 8 + 3 + 6.
(b) To turn in.
(c) To turn in.
(d)
Q8 = {1, i, j, k, −1, −i, −j, −k},
where i2 = j 2 = k 2 = ijk = −1. Let’s use the Cayley table to help us find the conjugacy classes:
×
1
i
j
k
−i
−j
−k
−1
1
1
i
j
k
−i
−j
−k
−1
i
i
−1
−k
j
1
k
−j
−i
j
j
k
−1
−i
−k
1
i
−j
k
k
−j
i
−1
j
−i
1
−k
−i
−1
1
k
−j
−1
−k
j
i
−j
−j
−k
1
i
k
−1
−i
j
−k
−k
j
−i
1
−j
i
−1
k
−1
−1
−i
−j
−k
i
j
k
1
Since the first row equals the first column 1 ∈ Z(Q8 ). Since the last row equals the last column, then
−1 ∈ Z(Q8 ). Every other row is not equal to its corresponding column, so the center contains just 1
and -1. Therefore
Z(Q8 ) = {1, −1}.
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Now let’s find the conjugacy class containing i. Let’s compute an example: jij −1 = −jij = −jk = −i,
so −i is in the conjugacy class of i. If we compute xix−1 for all x ∈ Q8 , we get the following set:
Oi = {i, −i}.
Since i, j, k are symmetric, then
Oj = {j, −j}
Ok = {k, −k}.
So the conjugacy classes are {i, −i}, {j, −j}, {k, −k} and the center is {1, −1}. The class equation
looks like
8 = 2 + 2 + 2 + 2.
Problem 6. (Exercise 20)
A group acts faithfully on a G-set X if the identity is the only element of G that leaves every element of
X fixed. Show that G acts faithfully on X if and only if no two distinct elements of G have the same action
on each element of X.
Solution 6. To turn in.
Problem 7. (Exercise 25)
If G is a group of order pn , where p is prime and n ≥ 2, show that G must have a proper subgroup of order
p. If n ≥ 3, is it true that G will have a proper subgroup of order p2 ?
Solution 7. Let g =
6 1 be an element of G. Then |g| =
6 1 and |g| | pn . Therefore |g| = pk for some positive
pk−1
integer k. Now, let h = g
. Then the order of h is
|h| = |g p
k−1
|=
pk
pk
|g|
=
= k−1 = p.
k−1
k
k−1
gcd (|g|, p
)
gcd (p , p
)
p
Therefore hhi is a subgroup of G with order p (and it is proper since it’s not the whole group).
Now if G is a group of order pn with n ≥ 3, then if there is any element g of order pk with k ≥ 2, there
k−2
exists an element with order p2 (by doing a similar construction as above, but this time letting h = g p ).
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This subgroup would also be proper since the order of the group is at least p . So the only way that G could
avoid a subgroup of order p2 is if every non-identity element of G has order p. Let’s consider this scenario
where we have every element in G with order p. The center of G has pt elements with t ≥ 1 by the class
equation. Therefore there exists an nonidentity h ∈ Z(G). Now let k 6∈ hhi. Since h and k have order p and
k 6∈ hhi, then hhi ∩ hki = {1}. Since h commmutes with everything, then if ha ∈ hhi and k b ∈ hki, then
ha k b = ha−1 (hk b ) = ha−1 k b h = ha−2 k b h2 = . . . = k b ha .
Therefore all the elements of hhi commute with all the elements of hki. Therefore hhihki is a subgroup of G
and it has order p2 . So if Z(G) = hhi, then G has a subgroup of order p2 .
Therefore there is a proper subgroup of order p2 in any group of order pn with n ≥ 3.
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